Answer :
Certainly! Let's differentiate the function [tex]\(y = \frac{2x}{5 - \tan(x)}\)[/tex] with respect to [tex]\(x\)[/tex]. Follow these steps:
### Step-by-Step Differentiation:
1. Identify the function:
[tex]\[ y = \frac{2x}{5 - \tan(x)} \][/tex]
2. Apply the Quotient Rule:
The Quotient Rule for differentiation states that if you have a function of the form [tex]\(\frac{u}{v}\)[/tex], where [tex]\(u\)[/tex] and [tex]\(v\)[/tex] are functions of [tex]\(x\)[/tex],
[tex]\[ \left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \][/tex]
Here, [tex]\(u = 2x\)[/tex] and [tex]\(v = 5 - \tan(x)\)[/tex].
3. Differentiate [tex]\(u\)[/tex] and [tex]\(v\)[/tex]:
- [tex]\(u = 2x\)[/tex], so [tex]\(u' = 2\)[/tex]
- [tex]\(v = 5 - \tan(x)\)[/tex], so [tex]\(v' = -\sec^2(x)\)[/tex] (since the derivative of [tex]\(\tan(x)\)[/tex] is [tex]\(\sec^2(x)\)[/tex])
4. Substitute [tex]\(u\)[/tex], [tex]\(u'\)[/tex], [tex]\(v\)[/tex], and [tex]\(v'\)[/tex] into the Quotient Rule:
[tex]\[ \frac{d}{dx} \left( \frac{2x}{5 - \tan(x)} \right) = \frac{(5 - \tan(x)) \cdot 2 - 2x \cdot (-\sec^2(x))}{(5 - \tan(x))^2} \][/tex]
5. Simplify the numerator:
[tex]\[ = \frac{2(5 - \tan(x)) + 2x \cdot \sec^2(x)}{(5 - \tan(x))^2} \][/tex]
6. Further simplify:
[tex]\[ = \frac{10 - 2\tan(x) + 2x \sec^2(x)}{(5 - \tan(x))^2} \][/tex]
7. Combine like terms:
[tex]\[ = \frac{2x \sec^2(x) + 10 - 2\tan(x)}{(5 - \tan(x))^2} \][/tex]
Thus, the derivative [tex]\(\frac{dy}{dx}\)[/tex] of the function [tex]\(y = \frac{2x}{5 - \tan(x)}\)[/tex] is:
[tex]\[ \boxed{\frac{2x \sec^2(x) + 10 - 2\tan(x)}{(5 - \tan(x))^2}} \][/tex]
### Step-by-Step Differentiation:
1. Identify the function:
[tex]\[ y = \frac{2x}{5 - \tan(x)} \][/tex]
2. Apply the Quotient Rule:
The Quotient Rule for differentiation states that if you have a function of the form [tex]\(\frac{u}{v}\)[/tex], where [tex]\(u\)[/tex] and [tex]\(v\)[/tex] are functions of [tex]\(x\)[/tex],
[tex]\[ \left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \][/tex]
Here, [tex]\(u = 2x\)[/tex] and [tex]\(v = 5 - \tan(x)\)[/tex].
3. Differentiate [tex]\(u\)[/tex] and [tex]\(v\)[/tex]:
- [tex]\(u = 2x\)[/tex], so [tex]\(u' = 2\)[/tex]
- [tex]\(v = 5 - \tan(x)\)[/tex], so [tex]\(v' = -\sec^2(x)\)[/tex] (since the derivative of [tex]\(\tan(x)\)[/tex] is [tex]\(\sec^2(x)\)[/tex])
4. Substitute [tex]\(u\)[/tex], [tex]\(u'\)[/tex], [tex]\(v\)[/tex], and [tex]\(v'\)[/tex] into the Quotient Rule:
[tex]\[ \frac{d}{dx} \left( \frac{2x}{5 - \tan(x)} \right) = \frac{(5 - \tan(x)) \cdot 2 - 2x \cdot (-\sec^2(x))}{(5 - \tan(x))^2} \][/tex]
5. Simplify the numerator:
[tex]\[ = \frac{2(5 - \tan(x)) + 2x \cdot \sec^2(x)}{(5 - \tan(x))^2} \][/tex]
6. Further simplify:
[tex]\[ = \frac{10 - 2\tan(x) + 2x \sec^2(x)}{(5 - \tan(x))^2} \][/tex]
7. Combine like terms:
[tex]\[ = \frac{2x \sec^2(x) + 10 - 2\tan(x)}{(5 - \tan(x))^2} \][/tex]
Thus, the derivative [tex]\(\frac{dy}{dx}\)[/tex] of the function [tex]\(y = \frac{2x}{5 - \tan(x)}\)[/tex] is:
[tex]\[ \boxed{\frac{2x \sec^2(x) + 10 - 2\tan(x)}{(5 - \tan(x))^2}} \][/tex]
